Quiz 2 - CMU Math
... Proof. Clearly W is nonempty. Then you may directly prove W is closed under addition and scalar multiplication. But the following method is more convenient. Since ...
... Proof. Clearly W is nonempty. Then you may directly prove W is closed under addition and scalar multiplication. But the following method is more convenient. Since ...
LOYOLA COLLEGE (AUTONOMOUS), CHENNAI – 600 034 1
... 11. If A is any nxm matrix such that AB and BA are both defined. Show that B is an mxn matrix. 12. If A is any square matrix, then show that A + A' is symmetric and A – A' is Skew - symmetric. 13. Prove that every invertible matrix possesses a unique inverse. ...
... 11. If A is any nxm matrix such that AB and BA are both defined. Show that B is an mxn matrix. 12. If A is any square matrix, then show that A + A' is symmetric and A – A' is Skew - symmetric. 13. Prove that every invertible matrix possesses a unique inverse. ...
computer science 349b handout #36
... converge if A is defective, and even if the dominant eigenvalue λ1 has multiplicity > 1. In this latter case, the method will converge with the same order of convergence as in the nondefective case if λ1 has a full set of eigenvectors (but very slowly if not). The critical limitation of the Power Me ...
... converge if A is defective, and even if the dominant eigenvalue λ1 has multiplicity > 1. In this latter case, the method will converge with the same order of convergence as in the nondefective case if λ1 has a full set of eigenvectors (but very slowly if not). The critical limitation of the Power Me ...
4. Transition Matrices for Markov Chains. Expectation Operators. Let
... 4. Transition Matrices for Markov Chains. Expectation Operators. Let us consider a system that at any given time can be in one of a finite number of states. We shall identify the states by {1, 2, . . . , N }. The state of the system at time n will be denoted by xn . The system is ’noisy’ so that xn ...
... 4. Transition Matrices for Markov Chains. Expectation Operators. Let us consider a system that at any given time can be in one of a finite number of states. We shall identify the states by {1, 2, . . . , N }. The state of the system at time n will be denoted by xn . The system is ’noisy’ so that xn ...
– Matrices in Maple – 1 Linear Algebra Package
... The result needs a little explaining. There are three items in the list. In each item are two scalars and a vector. The vector is the actual eigenvector. The scalars specify the eigenvalue and its multiplicity. In this case there are three distinct eigenvalues with multiplicity 1. The first in the l ...
... The result needs a little explaining. There are three items in the list. In each item are two scalars and a vector. The vector is the actual eigenvector. The scalars specify the eigenvalue and its multiplicity. In this case there are three distinct eigenvalues with multiplicity 1. The first in the l ...
PDF
... You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. ...
... You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. ...
Classification of linear transformations from R2 to R2 In mathematics
... Classification of linear transformations from R2 to R2 In mathematics, one way we “understand” mathematical objects is to classify them (when we can). For this, we have some definition of the objects as being isomorphic (essentially the same), and then understand when two objects are isomorphic. If ...
... Classification of linear transformations from R2 to R2 In mathematics, one way we “understand” mathematical objects is to classify them (when we can). For this, we have some definition of the objects as being isomorphic (essentially the same), and then understand when two objects are isomorphic. If ...
Set 3
... where V is a vector. Note that F (0) = V . Find the vector V and the matrix A that describe each of the following mappings [here the light blue F is mapped to the dark red F ]. ...
... where V is a vector. Note that F (0) = V . Find the vector V and the matrix A that describe each of the following mappings [here the light blue F is mapped to the dark red F ]. ...
UNIVERSITY OF OSLO Faculty of mathematics and natural sciences
... Let A, B, C, E ∈ Rn,n be matrices where AT = A. In this problem an (arithmetic) operation is an addition or a multiplication. We ask about exact numbers of operations. ...
... Let A, B, C, E ∈ Rn,n be matrices where AT = A. In this problem an (arithmetic) operation is an addition or a multiplication. We ask about exact numbers of operations. ...
Homework2-F14-LinearAlgebra.pdf
... [3] Find the 3 × 3 matrix which vanishes on the vector (1, 1, 0), and maps each point on the plane x + 2y + 2z = 0 to itself. [4] Find the 3 × 3 matrix that projects orthogonally onto the line ...
... [3] Find the 3 × 3 matrix which vanishes on the vector (1, 1, 0), and maps each point on the plane x + 2y + 2z = 0 to itself. [4] Find the 3 × 3 matrix that projects orthogonally onto the line ...
Differential Equations with Linear Algebra
... Suppose that the vectors u1 , u2 and u3 in a vector space V are linearly independent. Show that the vectors u1 + u2 , u2 + u3 and u3 + u1 are also linearly independent. 3.5. Diagonalizing Linear Maps. (20 points) The following matricesArepresent maps T : R3 → R3 with respect to the standard ...
... Suppose that the vectors u1 , u2 and u3 in a vector space V are linearly independent. Show that the vectors u1 + u2 , u2 + u3 and u3 + u1 are also linearly independent. 3.5. Diagonalizing Linear Maps. (20 points) The following matricesArepresent maps T : R3 → R3 with respect to the standard ...
Exam 3 Solutions
... b) Is A diagonalizable? If so, find a matrix P such that P −1 AP is diagonal, and display the diagonal matrix P −1 AP . Solution: The algebraic and geometric multiplicities are equal for each of the two eigenvalues, so A is diagonalizable. basis A ...
... b) Is A diagonalizable? If so, find a matrix P such that P −1 AP is diagonal, and display the diagonal matrix P −1 AP . Solution: The algebraic and geometric multiplicities are equal for each of the two eigenvalues, so A is diagonalizable. basis A ...
Review Sheet
... - For the right basis as the columns of P, A = PJP-1 - Generalized eigenvectors, generalized eigenspaces - Cycle of generalized eigenvectors -Similar matrices have the same Jordan canonical form ...
... - For the right basis as the columns of P, A = PJP-1 - Generalized eigenvectors, generalized eigenspaces - Cycle of generalized eigenvectors -Similar matrices have the same Jordan canonical form ...
University of Bahrain
... c) If A is n n matrix then (i) A diagonalizable only if A has n different eigenvalues. (ii) If 0 is an eigenvalue of A then A is not singular. ...
... c) If A is n n matrix then (i) A diagonalizable only if A has n different eigenvalues. (ii) If 0 is an eigenvalue of A then A is not singular. ...
Jordan normal form
In linear algebra, a Jordan normal form (often called Jordan canonical form)of a linear operator on a finite-dimensional vector space is an upper triangular matrix of a particular form called a Jordan matrix, representing the operator with respect to some basis. Such matrix has each non-zero off-diagonal entry equal to 1, immediately above the main diagonal (on the superdiagonal), and with identical diagonal entries to the left and below them. If the vector space is over a field K, then a basis with respect to which the matrix has the required form exists if and only if all eigenvalues of the matrix lie in K, or equivalently if the characteristic polynomial of the operator splits into linear factors over K. This condition is always satisfied if K is the field of complex numbers. The diagonal entries of the normal form are the eigenvalues of the operator, with the number of times each one occurs being given by its algebraic multiplicity.If the operator is originally given by a square matrix M, then its Jordan normal form is also called the Jordan normal form of M. Any square matrix has a Jordan normal form if the field of coefficients is extended to one containing all the eigenvalues of the matrix. In spite of its name, the normal form for a given M is not entirely unique, as it is a block diagonal matrix formed of Jordan blocks, the order of which is not fixed; it is conventional to group blocks for the same eigenvalue together, but no ordering is imposed among the eigenvalues, nor among the blocks for a given eigenvalue, although the latter could for instance be ordered by weakly decreasing size.The Jordan–Chevalley decomposition is particularly simple with respect to a basis for which the operator takes its Jordan normal form. The diagonal form for diagonalizable matrices, for instance normal matrices, is a special case of the Jordan normal form.The Jordan normal form is named after Camille Jordan.